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Talk:Array notation
Using the TM to bound \(\Sigma(n, c)\) How can we use the TM to bound \(\Sigma(n, c)\) vs. \(\{n, n (1) 2\}\)? FB100Z • talk • 22:18, May 22, 2013 (UTC) :Little Peng constructed TM that able to compute any Bowerian linear array with 109 states and 5 colors. With input x|111...111|111...111|...|111...111|111...111| (with n sections and n 1's in each) output will be \(\{n,n (1) 2\}\), and it requires \(n^2+n+2\) symbols. Thus, \(\Sigma(109+n^2+n+2,5) \gg \{n,n (1) 2\}\). Ikosarakt1 (talk ^ ) 22:39, May 22, 2013 (UTC) ::The correct syntax for the input is x|111...111 111...111 ... 111...111 111...111|, where each entries are divided by blanks. We could do even better, though; using the input x|111 11 1 1 1 ... 1 1 11| (with n+1 entries), the output will be \(\{3,n (1) 2\}\), and it requires \(2n+8\) symbols. Thus, \(\Sigma(109+2n+8,5) \gg \{3,n (1) 2\}\). ::But actually, the number of symbols printed by these TM's aren't actually large enough; we already know that even \(\Sigma(48,3)\) reaches the limit of tetrational arrays, let alone linear arrays. The TM used here is specialized for processing the "natural input" of linear arrays. -- I want more 03:02, May 23, 2013 (UTC) Some haphazard guesses about sizes of \(\Sigma(n)\) (may be wrong): \(\Sigma(6)\) can be expressed in terms of up-arrow notation with small number of arrows. \(\Sigma(7)\) surpasses Graham's number and can be bounded by \(\{3,n,1,2\}\) with quite large n. \(\Sigma(8)\) cannot be expressed using chained arrow notation with reasonable number of arrows. \(\Sigma(9)\) lies somewhere in linear array notation level: it can be upper bounded by ultatri or so. \(\Sigma(10)\) can be expressed using multirowed array notation (planar arrays). \(\Sigma(11)\) is at level of multidimensional array notation. \(\Sigma(12)\) is at level of hyperdimensional array notation. \(\Sigma(13)\) is at level of nested array notation. \(\Sigma(20)\) is larger than meameamealokkapoowa oompa. \(\Sigma(30)\) is larger than N in Bird's array notation. These busy beavers, of course, will have extremely unpreventable behavior. Ikosarakt1 (talk ^ ) 09:58, May 23, 2013 (UTC) How did you guess all that?Boboris02 (talk) 17:35, November 25, 2016 (UTC) 4th rule A comment was posted to Japanese wiki, which asks how the array notation can grow beyond 4 entries. He calculates that no matter how long the entry is, it is equal to {a,a,a}. \(\{a,b,1,\ldots,1,2\} = \{a,a,a,\ldots,a,1\} = \{a,a,a,\ldots,a\} = \ldots = \{a,n,1,2\} = \{a,a,a,1\} = \{a,a,a\}\) He says that the parenthesized part in the following equation, a should be b, not a, to make the function grow. \(\{a,b,1,\ldots,1,c,\ldots\} = \{a,a,1,\ldots,\{a,b-1,……1,c,……\}c-1,\ldots\}\) Is it correct? : \(\{a,b,1,1,\ldots,1,c,\ldots\} = \{a,a,a,a,\ldots,\{a,b-1,……1,c,……\}c-1,\ldots\}\), not \(\{a,a,1,1,\ldots,\{a,b-1,……1,c,……\}c-1,\ldots\}\). I think he/she misunderstood the rules. -- ☁ I want more ⛅ 16:11, February 10, 2014 (UTC) :: I see. There was also miscalculation in {a,n,1,2} = {a,a,a,1}. Correctly {a,n,1,2} = {a,a,{a,n-1,1,2},1}. Kyodaisuu (talk) 16:39, February 10, 2014 (UTC) I've always found rule 4 a bit cryptic. This may sound like a weird question, but as the headline states, I've found Rule 4 to be a bit cryptic. For instance, if an array is {a,b,1} (that is, three entries, and the third is equal to 1), what is done? Do you truncate the 1, as per rule 2? I'm assuming that this is the case, but I'm not sure. And another question, just for clarification about how rule 4 works, if you have an array {a,b,1,1,1,1,c}, what, exactly, is done? I'm the sort of person who runs off of examples like these. TheMostAwesomer (talk) 17:59, January 24, 2016 (UTC) :For {a,b,1}, yes, you truncate the 1, so {a,b,1} = {a,b} = a^b. :For {a,b,1,1,1,1,c}, if b = 1 you get a, if c = 1 you get a^b. Otherwise you apply rule 4 to get {a,a,a,a,a,{a,b-1,1,1,1,1,c},c-1}. That is, we subract 1 from the first entry after the second that is greater than 1, and then we replace the previous entry with the original array but with the second entry decremented by 1. All previous entries are replaced by a. Does that make sense? Deedlit11 (talk) 19:36, January 24, 2016 (UTC) ::Yes, that helps a lot. Thanks! TheMostAwesomer (talk) 20:25, January 24, 2016 (UTC) Bowers's array notation on Wikipedia See on Wikipedia. Until about 12 hours ago, the article STILL said that a{1}b = a+b, thought that a b = a{b}a, and did not even mention 5-entry arrays and beyond--I made the edits correcting all that. However, I still think this article is unfinished, so I just want to bring this to the attention of the wiki here. ArtismScrub (talk) 14:35, February 7, 2018 (UTC)